How Do You Evaluate This Limit Using a Definite Integral?

[Saitama]

Problem:

Evaluate:
$$\lim_{n\rightarrow \infty}\int_0^1 \frac{nx^{n-1}}{1+x}\,dx$$

Attempt:
I used the series expansion:
$$\frac{1}{1+x}=\sum_{r=0}^{\infty} (-1)^rx^r$$
From above, I got:
$$\lim_{n\rightarrow \infty} \sum_{r=0}^{\infty} \frac{(-1)^rn}{n+r}$$
But I don't see how to proceed from here.

Is there a way to solve this by elementary methods? I ask this because when I enter the above limit in W|A, it shows me weird things.

Any help is appreciated. Thanks!

 

[polygamma]

You could make the substitution $u = x^{n}$ and then justify bringing the limit inside of the integral. Doing so you'll see that the answer is $\frac{1}{2}$.

 

[Opalg]

Problem:

Evaluate:
$$\lim_{n\rightarrow \infty}\int_0^1 \frac{nx^{n-1}}{1+x}\,dx$$
Is there a way to solve this by elementary methods?

Integrate by parts: $$\int_0^1\frac{nx^{n-1}}{1+x}\,dx = \Bigl[\dfrac{x^n}{1+x}\Bigr]_0^1 + \int_0^1\dfrac{x^n}{(1+x)^2}\,dx.$$ But $\dfrac{x^n}{(1+x)^2} < x^n$ when $x > 0$, so that $$\int_0^1\dfrac{x^n}{(1+x)^2}\,dx < \int_0^1 x^n\,dx \to0$$ as $n\to\infty$. That will quickly lead to RV's answer $\frac12.$

 

[Saitama]

You could make the substitution $u = x^{n}$ and then justify bringing the limit inside of the integral. Doing so you'll see that the answer is $\frac{1}{2}$.

Ah, I never thought of it, thanks a lot Random Variable.

Integrate by parts: $$\int_0^1\frac{nx^{n-1}}{1+x}\,dx = \Bigl[\dfrac{x^n}{1+x}\Bigr]_0^1 + \int_0^1\dfrac{x^n}{(1+x)^2}\,dx.$$ But $\dfrac{x^n}{(1+x)^2} < x^n$ when $x > 0$, so that $$\int_0^1\dfrac{x^n}{(1+x)^2}\,dx < \int_0^1 x^n\,dx \to0$$ as $n\to\infty$. That will quickly lead to RV's answer $\frac12.$

This is a very nice solution, thank you Opalg! :)